Higher-Order Limits: Taylor Expansion & L'Hôpital Mastery

Introduction

When standard limit formulas hit a wall—typically for $\frac{0}{0}$ forms where the leading terms cancel—you need to go to the next order. Taylor (Maclaurin) series expansion and L'Hôpital's rule are your two power tools for this. JEE Main features 20–25 such "higher-order" limit problems over the past six years. This article teaches you when to use which tool and how to execute both flawlessly.

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1. Essential Taylor (Maclaurin) Series

Memorize these expansions up to the order shown—this is sufficient for 99% of JEE problems.

The Big Six

$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots$

$\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots$

$\tan x = x + \frac{x^3}{3} + \frac{2x^5}{15} + \cdots$

$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$

$\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots \quad (|x| \leq 1, x \neq -1)$

$(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \cdots \quad (|x| < 1)$

Derived Series (Used Frequently)

$\sin^{-1} x = x + \frac{x^3}{6} + \cdots, \qquad \tan^{-1} x = x - \frac{x^3}{3} + \cdots$

$\sinh x = x + \frac{x^3}{3!} + \cdots, \qquad \cosh x = 1 + \frac{x^2}{2!} + \cdots$

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2. The Taylor Series Method for Limits

When to Use

Use Taylor expansion when:

• The limit has the form $\frac{0}{0}$ after standard limit results fail.
• The numerator and denominator have matching leading terms that cancel, and you need the next-order term.
• L'Hôpital would require 2+ rounds of differentiation (Taylor is faster).

The Technique

Step 1: Expand each function in the expression up to the required order.

Step 2: Substitute the expansions and simplify (combine like terms).

Step 3: Cancel the lowest-order terms (they give $0/0$).

Step 4: The ratio of the next surviving terms gives the limit.

Determining the Required Order

The order needed equals the power of $x$ in the denominator. If the denominator is $x^3$, expand everything to at least the $x^3$ term.

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3. L'Hôpital's Rule: Precise Statement

The Rule

If $\lim_{x \to a} f(x) = \lim_{x \to a} g(x) = 0$ (or both $\pm\infty$), and $g'(x) \neq 0$ near $a$, then: $\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}$ provided the right-hand limit exists (or is $\pm\infty$).

When to Use

Use L'Hôpital when:

• The expression naturally has a simple derivative.
• Only one application is needed.
• Taylor expansions would be messier (rare, but happens with compositions).

When NOT to Use

• When the limit is not in $\frac{0}{0}$ or $\frac{\infty}{\infty}$ form.
• When standard limits or algebraic manipulation work faster.
• When it leads to a circular application (e.g., proving $\lim \frac{\sin x}{x} = 1$ using L'Hôpital requires knowing $(\sin x)' = \cos x$, which itself depends on this limit).

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4. Worked Examples

Example 1 (Taylor — Second-Order Cancellation)

Evaluate $\displaystyle \lim_{x \to 0} \frac{e^x - 1 - x}{x^2}$.

Solution (Taylor):

Denominator is $x^2$, so expand $e^x$ to the $x^2$ term: $e^x = 1 + x + \frac{x^2}{2} + \cdots$ $\frac{e^x - 1 - x}{x^2} = \frac{\frac{x^2}{2} + O(x^3)}{x^2} = \frac{1}{2}.$

Solution (L'Hôpital): Apply twice: $\frac{e^x - 1}{2x} \to \frac{e^x}{2} = \frac{1}{2}.$

Both methods give $\frac{1}{2}$, but Taylor was faster (one step vs. two).

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Example 2 (Taylor — Third-Order)

Evaluate $\displaystyle \lim_{x \to 0} \frac{\sin x - x + \frac{x^3}{6}}{x^5}$.

Solution:

Expand $\sin x$ to the $x^5$ term: $\sin x = x - \frac{x^3}{6} + \frac{x^5}{120} - \cdots$ $\sin x - x + \frac{x^3}{6} = \frac{x^5}{120} - \cdots$ $\frac{\frac{x^5}{120}}{x^5} = \frac{1}{120}.$

Using L'Hôpital here would require 5 rounds of differentiation—clearly impractical.

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Example 3 (Taylor — Trig Combination)

Evaluate $\displaystyle \lim_{x \to 0} \frac{\tan x - \sin x}{x^3}$.

Solution:

$\tan x - \sin x = \left(x + \frac{x^3}{3} + \cdots\right) - \left(x - \frac{x^3}{6} + \cdots\right) = \frac{x^3}{3} + \frac{x^3}{6} + \cdots = \frac{x^3}{2} + \cdots$ $\lim = \frac{1}{2}.$

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Example 4 (L'Hôpital — Clean Single Application)

Evaluate $\displaystyle \lim_{x \to 0} \frac{x - \sin x}{x^2 \tan x}$.

Solution:

This is $\frac{0}{0}$. However, let's simplify first: $\frac{x - \sin x}{x^2 \tan x} = \frac{x - \sin x}{x^3} \cdot \frac{x}{\tan x}.$

$\frac{x}{\tan x} \to 1$. For $\frac{x - \sin x}{x^3}$, use Taylor: $x - \sin x = \frac{x^3}{6} + \cdots \Rightarrow \frac{x - \sin x}{x^3} \to \frac{1}{6}.$

Answer: $\frac{1}{6}$.

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Example 5 (Taylor — Logarithmic)

Evaluate $\displaystyle \lim_{x \to 0} \frac{x - \ln(1+x)}{x^2}$.

Solution:

$\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots$ $x - \ln(1+x) = \frac{x^2}{2} - \frac{x^3}{3} + \cdots$ $\frac{x - \ln(1+x)}{x^2} \to \frac{1}{2}.$

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Example 6 (Taylor — Two-Variable Cancellation)

Evaluate $\displaystyle \lim_{x \to 0} \frac{e^x - e^{\sin x}}{x - \sin x}$.

Solution:

Let $u = x - \sin x \to 0$. Write: $e^x - e^{\sin x} = e^{\sin x}(e^{x - \sin x} - 1).$ $\frac{e^{\sin x}(e^u - 1)}{u} \to e^0 \cdot 1 = 1.$

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5. Previous Year JEE Problems

Problem 1 (JEE Main 2023)

$\displaystyle \lim_{x \to 0} \frac{e^x \sin x - x - x^2}{x^3} =$ ?

Solution:

Expand: $e^x \sin x = \left(1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots\right)\left(x - \frac{x^3}{6} + \cdots\right).$ Multiply term by term (keeping up to $x^3$): $= x + x^2 + \frac{x^3}{2} - \frac{x^3}{6} + \frac{x^3}{6} + \cdots = x + x^2 + \frac{x^3}{2} - \frac{x^3}{6} + \frac{x^3}{6} + \cdots$

Let me be more careful: $e^x \sin x = 1 \cdot x + x \cdot x + \frac{x^2}{2} \cdot x + 1 \cdot \left(-\frac{x^3}{6}\right) + \cdots$ $= x + x^2 + \frac{x^3}{2} - \frac{x^3}{6} + \cdots = x + x^2 + \frac{x^3}{3} + \cdots$

So: $\frac{e^x \sin x - x - x^2}{x^3} = \frac{\frac{x^3}{3} + \cdots}{x^3} \to \frac{1}{3}.$

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Problem 2 (JEE Main 2021)

$\displaystyle \lim_{x \to 0} \frac{x(\cos x - 1)}{x - \sin x} =$ ?

Solution:

Numerator: $x(\cos x - 1) = x\left(-\frac{x^2}{2} + \frac{x^4}{24} - \cdots\right) = -\frac{x^3}{2} + O(x^5)$.

Denominator: $x - \sin x = \frac{x^3}{6} - O(x^5)$.

$\frac{-x^3/2}{x^3/6} = \frac{-1/2}{1/6} = -3.$

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Problem 3 (JEE Main 2024)

$\displaystyle \lim_{x \to 0} \frac{x^2 - \sin^2 x}{x^2 \sin^2 x} =$ ?

Solution:

$\sin^2 x = x^2 - \frac{x^4}{3} + \cdots$ (from $\sin x = x - x^3/6 + \cdots$, squaring).

Numerator: $x^2 - \sin^2 x = x^2 - x^2 + \frac{x^4}{3} - \cdots = \frac{x^4}{3} + \cdots$.

Denominator: $x^2 \sin^2 x = x^2 \cdot (x^2 - \frac{x^4}{3} + \cdots) = x^4 - \cdots$.

$\frac{x^4/3}{x^4} = \frac{1}{3}.$

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6. Taylor vs. L'Hôpital: Decision Guide

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7. Common Higher-Order Limit Results (Memorize!)

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8. Tips for JEE Aspirants

1. Taylor first, L'Hôpital second. For JEE, Taylor expansion is almost always the faster and safer choice.
2. Match the order to the denominator. If the denominator is $x^k$, expand everything to the $x^k$ term.
3. Memorize the Big Six expansions to at least the $x^3$ term. For $\sin x$ and $e^x$, go to $x^5$ and $x^4$ respectively.
4. Product of series: multiply term by term and collect powers. Don't expand fully—stop at the required order.
5. Don't apply L'Hôpital blindly. Check that the form is indeed $\frac{0}{0}$ or $\frac{\infty}{\infty}$ before and after each application.
6. Cross-check with standard results. If you know $\frac{\sin x - x}{x^3} = -\frac{1}{6}$, use it directly—no need to re-derive.

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9. Quick-Reference Summary

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The Taylor expansion method is arguably the single most powerful technique for JEE limits. It converts every higher-order indeterminate form into a routine algebraic simplification. Invest time in memorizing the key series—it pays dividends across Limits, Differentiability, and even Definite Integration.